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meera_yadav Group Title

How do we derive the expression for the potential due to a point charge at a distance r away from it (taking potential at infinity to be zero) ?

  • 2 years ago
  • 2 years ago

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  1. quarkine Group Title
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    integrate the work done in bringing a unit charge from infinity to that point at distance r using coulomb's law

    • 2 years ago
  2. meera_yadav Group Title
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    thanks quakine, but this is where i'm struck I take Q due to which i need to calculate potential at point A I bring a test charge from infinity to the point A without any acceleration ! we have \[w = \int\limits_{\infty}^{A}F.dl\] Here F is the external force that an external system applies to bring the test charge to A let the test charge be a unit positive test charge. the force F is equal and opposite to that of the force due to Q therefore I substitute F= -E { E is the electric field due to Q } therefore we've \[w = -\int\limits_{\infty}^{A}E.dl\] now I substitute for E due to Q , i.e\[E=Q/4\Pi \in0 r ^{2} \] direction of E is opposite to the displacement dl ! here's i'm struck !

    • 2 years ago
  3. CarlosGP Group Title
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    Just solve the integral and substitute the extreme values, taking into account that the value in infinite is zero. Some clue: the integral function of (1/r^2)dr is -1/r

    • 2 years ago
  4. quarkine Group Title
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    there are two forces in action here,the electrostatic repulsion and your force which overcomes this repulsion and bring these two charges closer.due to the no acc condition both forces are equal and opposite. |dw:1246119205782:dw| F is the force you apply against F1 which is the electrostratic repulsion. hence F= -F1 as these two forces are opposite in direction but equal. Also, it is hard to know how much force you applied but it is easier to measure the repulsion.so F in the integral F.dl is Your force but we replace it by a minus F1. so the direction of dl and YOUR force are always same (towards Q) and work is positive. the minus before the integral of E.dl already takes care of sign (cos180 is -1)|dw:1246120234592:dw|. so you can forget about the minus sign and evaluate the integral as drawn.

    • 2 years ago
  5. Vincent-Lyon.Fr Group Title
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    \(\vec E=-\vec \nabla V= - \vec {grad}\:V\) So \(V(r)=-\int E(r) \,dr \) Take the primitive of KQ/r², change sing, then chose constant to be zero at infinity. The advantage of this method is that you do not need to refer to energy, neither to imagine a force counteracting the electrostatic force and working out the work done by this force.

    • 2 years ago
  6. meera_yadav Group Title
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    thanks all :) quarkine, if I solve

    • 2 years ago
  7. meera_yadav Group Title
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    |dw:1340944945729:dw| i get finally \[-kQ/ra\] this is negative of what the answer is !

    • 2 years ago
  8. meera_yadav Group Title
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    answer is KQ/r

    • 2 years ago
  9. Vincent-Lyon.Fr Group Title
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    Correct !

    • 2 years ago
  10. meera_yadav Group Title
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    its coming out after solving integral (as can be seen above ) -kQ/r . let me know what's the flaw in the way i solved.

    • 2 years ago
  11. quarkine Group Title
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    the minus sign is fine...it gets away when we exchange the upper and lower limit.. [it's a general property of definite integral] if u r nt familiar with that then just evaluate your integral ans with proper lmits..you will get the correct answer

    • 2 years ago
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